@Preben Alsholm I run this now and it seems to work, although the graph from 

animate(plot,['p(aa,P,[F,F1])',0..aa,-K..1],P=-.7..-.06);

gives a result more like i am expecting than the final solution. 

odeplot(res,[[x,f(x)],[x,diff(f(x),x)]],0..sol[1]);

Also, I think P should be positive although when i try to impliment this it really confuses itself (or me!), and what is the final value for P? Does it produce this? 

Finally, where have you got the numbers 0.7, -0.6 and -0.06 from?

@Preben Alsholm I have managed to manipulate the code slightly to get a solution i am reasonably happy with, but could you explain where you the rhs from in the line: 

fsolve([p1,p2,p3,p4],[1.5,-1.5,1,1]);

so where do the numbers 1.5, -1.5, 1,1 come from. I could a solution from this by guessing at them but just because it looks right doesnt mean i have it correct.

@Preben Alsholm i think if you use both the condition above you only need one, the one for positive a (f(a)=0) is 

-(3)^(-1/2), that is minus root three over three. 

@Preben Alsholm The ODE is:

where e=0.1, and alpha = Pi/3. The parameter is P. The other boundary conditions I am using other than the one in question are D(h)(0)=0, and h(0)=1.

in text the ode is:

-diff(f(x),x,x)/((1+diff(f(x),x)^2)^(3/2)) = e*(x*sin(alpha)+f(x)*cos(alpha))+P.

@Preben Alsholm  yes the values will be different, and its not as simply as K=-k either. Although its pretty close. Also im not certain f only has 2 zeros over the entire domain, but between -b and a it has none.

There should only be two points with f(x)=0, one positive and one negative. And yes that is what I mean but I dont want to fix the point a. My idea is to solve the BVP from 0 to a and then again from -b to 0 (where f(-b)=0) and combine these solutions but I cant find a way of finding this point a.